Question

4. (12 points ) Solve:

Answer 1

4. Let A=10 l and flt) = 1 1-1 od 1. The characteristic polynomial of A is. A() = 72-(faceA) + delta = y2 +1: 5 4(2)=0 + 1 = Iš Eigenvector of A Corresponding to = (A+ė) V = 0 5 11117_107 -1 ė] [ 2] O Re R2-8 hj .. ſ § Ilnl lo o LV2] "O.. vit VO év + V2 =0 I fet v=-1 then &q=a. -16-è fint)] et v = "" - Lille-i ton) I sinth fet se (t) = [ - Goti tes y Ha.. sont I lost | 41H and Xa(t) are linearly independent solution of x = Az. 1. X 14=4141424 testo = q elmi

fet fint - lost P=1 I first I-lost fint forst] Got Cost 4 - sl-t host t font 2 t fint tlost] 1 -list of + S fint at | Istrist att fast at st Costot = t bint -5 bin tot using iste goupion by beats = t toint + list It bintat =-tlot + f fost at - = -tastt fint i [p'fl Hdp = 1 intfint - ast - Cost] -t los tt fint + fint] 1-toint-2 ast] -t lost +2 fint it *p1# = p. Sptfight - - I-Got föst] -t fint– 2 6st7 boint Gost] [-t Cost +2 tint] I tlostint+266²2 -t histlest ta bight: 7 | -tbinht-2 binttost -t let +2 fist fint I 27 1 2p (t) = The general solution of the given system is. 91 fint ? Gost] |-+]

2103 [3 - 17-449-89 = 11 => 1942] -1] => [= ? and.q=4 listy +2 hent- *1 = 12wt + bont -t). - G+27 i-last 1+2 i 2 sint-lost +27 12 lost t fint -t). ai 2(t) = 1 dint'. Las ast